Problem: Suppose we have a vector field $f(x, y) = \left( \dfrac{x}{y}, \dfrac{y}{x} \right)$ and a curve $C$ that is parameterized by $\alpha(t) = (\cos(t), \sin(t))$ for $0 < t < \dfrac{\pi}{2}$. What is the line integral of $f$ along $C$ ? $ \int_C f \cdot d\alpha = $
Solution: Given a vector field $f$, a parameterization $\alpha$, and bounds $t_0$ and $t_1$, we can calculate the line integral as follows: $ \int_C f \cdot d\alpha = \int_{t_1}^{t_2} f(\alpha(t)) \cdot \alpha'(t) \, dt$ Here, $f(x, y) = \left( \dfrac{x}{y}, \dfrac{y}{x} \right)$ and $\alpha(t) = (\cos(t), \sin(t))$. $\begin{aligned} &f(\alpha(t)) = \left( \dfrac{\cos(t)}{\sin(t)}, \dfrac{\sin(t)}{\cos(t)} \right) \\ \\ &\alpha'(t) = (-\sin(t), \cos(t)) \end{aligned}$ Now we can rewrite our line integral as a single-variable integral. $ \int_C f \cdot d\alpha = \int_0^{\frac{\pi}{2}} \left( \dfrac{\cos(t)}{\sin(t)}, \dfrac{\sin(t)}{\cos(t)} \right) \cdot (-\sin(t), \cos(t)) \, dt$ Let's solve the integral. $\begin{aligned} &\int_0^{\frac{\pi}{2}} \left( \dfrac{\cos(t)}{\sin(t)}, \dfrac{\sin(t)}{\cos(t)} \right) \cdot (-\sin(t), \cos(t)) \, dt \\ \\ &= \int_0^{\frac{\pi}{2}} -\cos(t) + \sin(t) \, dt \\ \\ &= -\left[ \sin(t) + \cos(t) \right]_0^{\frac{\pi}{2}} \\ \\ &= - (1 + 0 - (0 + 1)) \\ \\ &= 0 \end{aligned}$ In conclusion, the line integral $ \int_C f \cdot d\alpha = 0$.